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Source: — Data Sufficiency |
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by LazySammy » Tue Feb 08, 2011 7:10 pm
I think the answer is C. Both statements together are sufficient. The area of of Equilateral triangle is S^2 sqrt(3) /4.

Given ABD is 60 tells you that this is an equilateral triangle.

Given AC = 12 helps you calculate area using the formula.

This is my best explanation. Let me know if this is right answer.
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by Night reader » Tue Feb 08, 2011 7:57 pm
LazySammy wrote:I think the answer is C. Both statements together are sufficient. The area of of Equilateral triangle is S^2 sqrt(3) /4.

Given ABD is 60 tells you that this is an equilateral triangle.

Given AC = 12 helps you calculate area using the formula.

This is my best explanation. Let me know if this is right answer.
In statement (1) ABD=60 tells you only ABD=60, how can you know that ACD is 60, 70 or even ... :) it can be anything
With statement (1) one should get information only about ABD=60, AD=6*SQRT(3),and the complement angle to ADB, in the triangle ABC, ADC=90. The unknown data are AC=?, DC=?, ACD=?, DAC=?

In statement (2) one knows only AC=12, the unknown data are BC=?, AB=?, BAD=?, ABD=?

How you came to know what was unknown in statements (1) and (2) with your reasoning about equilateral triangles
Night reader wrote:If AD is 6*SQRT(3), and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12
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by LazySammy » Tue Feb 08, 2011 8:16 pm
Given ADC = 90 implies ADB = 90 which further means DAB = 30

If DAB = 30 , DAC = 30. thus BAC = 60


Thus leaves us with C = 60.

That's is why I reasoned this to be a equilateral triangle.
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by rohu27 » Tue Feb 08, 2011 8:24 pm
1)angle ABD=60, so triangle ABD is a 30-60-90. we can find all the sides.(1:sqrt3:2) but we still dnt knw DC. so insuff.
2) AC=12, we can find DC

using 1 & 2 area = 1/2 bh.

so answer is C

Night reader wrote:If AD is 6*SQRT(3), and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12
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by LazySammy » Tue Feb 08, 2011 8:47 pm
A. Since Angle ABD = 60 we can imply that ABC is an equilateral triangle because Angle ADB = 90 , Angle BAD = 30
and thus DAB is also equal to 30

We know that AD = 6 sqrt (3) and because ADB is a 30-60-90 Triangle the sides are in the ratio of 1 : sqrt (3) : 2 using this we can tell that AB = 12 , BD = 6 and AD = 6 sqrt (3) (given)

In an equilateral triangle since all sides are equal AB = AC = BC = 12 and given BD = 6 , DC = 6. With these dimensions and AD = 6 sqrt (3) we can find the area of the triangle. Sufficient.

B. AC = 12 and AD = 6 SQRT (3) we can find out DC = 6. Now working backwards since the ratios of the sides are in the ration of 1:sqrt 3 : 2 we can say Angle C = 60 and still be able to derive the area by using logic from A above.
Answer is D. Each stmt alone is sufficient.


Let me know what you think.
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by rohu27 » Tue Feb 08, 2011 8:57 pm
i thnk u mean to say DAC is 30? but no whr does the question say AD is bisects A equally. its just a line drawn so tt it makes 90 wth the base. so we cant assume DAC to be 30 too.

hope im nt missing naythng here.
LazySammy wrote:A. Since Angle ABD = 60 we can imply that ABC is an equilateral triangle because Angle ADB = 90 , Angle BAD = 30
and thus DAB is also equal to 30

We know that AD = 6 sqrt (3) and because ADB is a 30-60-90 Triangle the sides are in the ratio of 1 : sqrt (3) : 2 using this we can tell that AB = 12 , BD = 6 and AD = 6 sqrt (3) (given)

In an equilateral triangle since all sides are equal AB = AC = BC = 12 and given BD = 6 , DC = 6. With these dimensions and AD = 6 sqrt (3) we can find the area of the triangle. Sufficient.

B. AC = 12 and AD = 6 SQRT (3) we can find out DC = 6. Now working backwards since the ratios of the sides are in the ration of 1:sqrt 3 : 2 we can say Angle C = 60 and still be able to derive the area by using logic from A above.
Answer is D. Each stmt alone is sufficient.


Let me know what you think.
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by Night reader » Tue Feb 08, 2011 9:06 pm
@sammy, I was drawing an additional figure for you to show the possibility of angles BAD and DAC to be different. The problem doesn't say anything about the angle DAC. We know only about AD, angle ADC and angle BAD. Angle DAC can be found in the interval {1...89} :)

why you like so equilateral triangles, see rohu's explanation
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by Night reader » Tue Feb 08, 2011 9:08 pm
st(1) we can find only two sides AB and BD
rohu27 wrote:1)angle ABD=60, so triangle ABD is a 30-60-90. we can find all the sides.(1:sqrt3:2) but we still dnt knw DC. so insuff.
2) AC=12, we can find DC

using 1 & 2 area = 1/2 bh.

so answer is C

Night reader wrote:If AD is 6*SQRT(3), and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12
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by ankur.agrawal » Tue Feb 08, 2011 9:11 pm
Night reader wrote:If AD is 6*SQRT(3), and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12
Pls post the OA.[/b]
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by Night reader » Tue Feb 08, 2011 9:56 pm
ankur.agrawal wrote:
Night reader wrote:If AD is 6*SQRT(3), and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12
Pls post the OA.[/b]
ops, I just went through CAT Mgmat haven't seen OA yet
I think it should be C
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by rohu27 » Tue Feb 08, 2011 10:32 pm
exactly. using st 2 we can find DC too.
so both the statemtns are needed nd C should be the answer.

Night reader wrote:st(1) we can find only two sides AB and BD
rohu27 wrote:1)angle ABD=60, so triangle ABD is a 30-60-90. we can find all the sides.(1:sqrt3:2) but we still dnt knw DC. so insuff.
2) AC=12, we can find DC

using 1 & 2 area = 1/2 bh.

so answer is C

Night reader wrote:If AD is 6*SQRT(3), and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

(2) AC = 12
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by Geva@EconomistGMAT » Wed Feb 09, 2011 12:38 am
In DS questions, we cannot trust the figure - only the data stated in the question stem and statements counts.
The triangle looks like an equilateral triangle, but this fact is not mentioned in the question stem. If the statement(s) somehow fix the triangle as equilateral, then we can find the area - using the 30:60:90 1:√3:2 ratio, we can find the side of the triangle from the height, and from that find the area. The issue is therefore whether the statement(s) are enough to prove that ABC is equilateral.

With this in mind, the answer is indeed C: each of the statements deals with one of the "halves" of big triangle ABC, but you need both in order to show that both halves are 30:60:90 triangles.

Stat. (1): If ABD= 60 degrees, then the left triangle is a 30:60:90 triangle. But as nightreader's later post shows, this does not tell us that the right half is also a 40:60:90 triangle, since we do not know that the height AD is also a bisector.

Stat (2): from AC=12 and the height, we can find DC using the pythagorean theorem. Guess what, it'll come down to 6. The right triangle ADC satisfies the 6:6√3:6*2 ratio of a 30:60:90 triangle, and we can therefore infer that its angles are indeed 30, 60 and 90, respectively. This takes care of the right triangle, but alone is insufficient, as it says nothing of the left side.

Combined: each of the triangle "halves" is a 30:60:90 triangle, and both share the same height AD=6√3. Therefore, the two trinagles are equal, the triangle ABC is indeed an equilateral triangle, and the area can be found using the equilateral formula side^2*√3/4 = 12√3/4 = 3√3. Sufficient.
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by vaflaly » Wed Feb 09, 2011 3:25 am
To solve the question, we must assume that C, D, B are in the same line. Otherwise, the answer is E
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by Night reader » Wed Feb 09, 2011 4:17 am
vaflaly wrote:To solve the question, we must assume that C, D, B are in the same line. Otherwise, the answer is E
irrelevant ---> angle ADC can be 90` when it is made by the line segments drawn from the points A-D-C as perpendiculars. The line segments should cross, and the line segment BC is placed on a cross-point.

ABC is a triangular region means that AB, AC and BC are line segments - and if BC is the line segment - the points B and C must be in the same line.
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